The following are excerpts from Folland:
Proposition 2.12:
2.12 Proposition. Let $(X, \mathcal{M}, \mu)$ be a measure space and let $(X, \bar{\mathcal{M}}, \bar{\mu})$ be its completion. If $f$ is an $\mathcal{M}$-measurable function on $X$, there is an $\bar{\mathcal{M}}$-measurable function $g$ such that $f = g$ $\bar{\mu}$-almost everywhere.
Proposition 2.23:
If $f, g \in L^1$, then $\int_E f = \int_E g$ for all $E \in \mathcal{M}$ iff $\int |f – g| = 0$ iff $f = g$ a.e.
This proposition shows that for the purposes of integration, it makes no difference if we alter functions on null sets. Indeed, one can integrate functions $f$ that are only defined on a measurable set $E$ whose complement is null simply by defining $f$ to be zero (or anything else) on $E^c$. In this fashion, we can treat $\bar{\mathbb{R}}$-valued functions that are finite a.e. as real-valued functions for the purposes of integration.
With this in mind, we shall find it more convenient to redefine $L^1(\mu)$ to be the set of equivalence classes of a.e.-defined integrable functions on $X$, where $f$ and $g$ are considered equivalent iff $f = g$ a.e. This new $L^1(\mu)$ is still a complex vector space (under pointwise a.e. addition and scalar multiplication). Although we shall henceforth view $L^1(\mu)$ as a space of equivalence classes, we shall still employ the notation "$f \in L^1(\mu)$" to mean that $f$ is an a.e.-defined integrable function. This minor abuse of notation is commonly accepted and rarely causes any confusion. The new definition of $L^1(\mu)$ has two further advantages. First, if $\bar{\mu}$ is the completion of $\mu$, Proposition 2.12 yields a natural one-to-one correspondence between $L^1(\bar{\mu})$ and $L^1(\mu)$, so we can (and shall) identify these spaces. Second, $L^1$ is a metric space with distance function $\rho(f,g) = \int |f – g|$. (The triangle inequality is easily verified, and obviously $\rho(f,g) = \rho(g,f)$; but to obtain the condition that $\rho(f,g) = 0$ only when $f = g$, one must identify functions that are equal a.e., according to Proposition 2.23b.) We shall refer to convergence with respect to this metric as convergence in $L^1$; thus $f_n \to f$ in $L^1$ iff $\int |f_n – f| \to 0$.
I'm confused about the following aspects:
- Did folland somehow assume that the measure is complete? Folland first shows how we can integrate functions defined on a measurable set $E$ whose complement is null. But then he talks about integration on a.e. defined functions. If the measure is not complete, then an a.e defined function is not necessarily defined on a measurable set.
- Folland mentions that $L^1(\mu)$ is the equivalence class of a.e. defined integrable functions. But what is the definition of integrability of a function is defined a.e instead of everywhere?
Then Folland proceeds to prove the dominated convergence theorem:
2.24 The Dominated Convergence Theorem. Let $ \{f_n\} $ be a sequence in $ L^1 $ such that (a) $ f_n \rightarrow f $ a.e., and (b) there exists a nonnegative $ g \in L^1 $ such that $ |f_n| \leq g $ a.e. for all n. Then $ f \in L^1 $ and $ f = \lim_{n \to \infty} f_n $.
Proof. $ f $ is measurable (perhaps after redefinition on a null set) by Propositions 2.11 and 2.12, and since $ |f| \leq g $ a.e., we have $ f \in L^1 $…
I'm confused how why you can redefine it on a null set. Of course, proposition 2.12 shows that after a redefinition $f$ is measurable, but why can you do that in the first place? Even if we view $L^1$ as a set of equivalence classes, it is only an equivalence class on measurable functions.
Best Answer
I'm not sure if Folland did assume (implicitly or explicitly) that the measure space it complete, but based on the presence of Proposition $2.12$, I'm guessing he did (although he may not have said to explicitly). Other things you're quoting here seem to confirm that. However, let's cover why, regardless of what Folland assumes or explicitly says, all of what he's doing here is "allowed" and "makes no difference."
Let $N$ denote the set of all $A\subset X$ such that there exists $E\in \mathcal{M}$ with $A\subset E$ and $\mu(E)=0$. We refer to the members of $N$ as null sets (so "null" doesn't imply measurable according to our usage).
Let $D$ denote the set of $\overline{\mathbb{R}}$-valued functions defined on a subset of $X$. For $f\in D$, let $\text{dom}(f)$ denote the set of $x\in X$ such that $f(x)$ is defined and let $\text{fin}(f)$ denote the set of points $x\in \text{dom}(f)$ such that $f(x)\in \mathbb{R}$.
Let $M_0$ denote the set of functions $f\in D$ which are defined on all of $X$, finite on all of $X$, and $\mathcal{M}$-measurable.
For $f\in D$ and $g\in M_0$, let $\text{bad}(f,g)$ denote the set of points $x\in X$ at which $f(x)$ is undefined, defined but infinite, or defined, finite, but not equal to $g(x)$. Let $M_1$ denote the set of $f\in D$ such that there exists $g\in M_0$ such that $\text{bad}(f,g)\in N$.
The set $M_0$ is the set where everything is unambiguous. The set $M_1$ is the set where we really want to build a theory of integration. Replacing $(X,\overline{M},\overline{\mu})$ gets us partway from $M_0$ to $M_1$, but not all the way there. How far it gets you from $M_0$ to $M_1$ depends on whether you allow functions that are everywhere defined but sometimes $\pm \infty$ to be considered measurable. I allow that, but Folland doesn't.
Whether we allow a function that's everywhere defined but not everywhere finite to be considered measurable isn't an obvious choice. Ultimately, it doesn't matter, because we want redefinition on a set of measure zero not to change our equivalence class in our theory of integration. But if you don't allow $\pm \infty$-valued functions to be considered measurable, then it's possible that the pointwise limit of a sequence of measurable functions is not measurable, because it is $\pm \infty$ at some points. If you do allow $\pm \infty$-valued functions to be considered measurable, then $f_1,f_2$ could both be measurable but $f_1+f_2$ is not everywhere defined. For example, it won't be defined on $\{x\in X:f_1(x)=+\infty\}\cap \{x\in X:f_2(x)=-\infty\}$, since $\infty-\infty$ is not defined. You can get around this with the convention that $\infty-\infty=0$, but we're getting around all of these "hiccups" with conventions, and our other conventions will make that one unnecessary.
Suppose we define a perfectly good theory of integration on $M_0$. That is, $\int fd\mu$ is defined for all $f\in M_0^+$, which is the set of $f\in M_0$ such that $f(x)\geqslant 0$ for all $x\in X$. For $f\in M_0$, $f^+:=\max\{0,f\}$ and $f^-=\max\{0,-f\}$ are members of $M_0^+$.
Define $H^+$ to be the set of all $f\in M_0^+$ such that $\int fd\mu<\infty$ and let $H$ denote the set of $f\in M_0$ such that $f^+,f^-\in H$. Then we can define $\int fd\mu=\int f^+ d\mu - \int f^-d\mu$. This defines integrals in the most restricted setting. Here, everything is unambiguous. We aren't required to define/redefine any functions anywhere.
Define an equivalence relation $\sim$ on $M_1$ by letting $f_1\sim f_2$ if there exists $g\in M_0$ such that $\text{bad}(f_1,g), \text{bad}(f_2,g)\in N$. Let $L_1$ denote the set of $\sim$-equivalence classes $C$ in $M_1$ such that $H\cap C\neq \varnothing$. For each $f\in C$, define $\int fd\mu=\int gd\mu$, where $g\in H\cap C$. One can show that this definition is independent of the choice of $g\in H\cap C$. That's because if $g_1,g_2\in H\cap C$, $g_1-g_2$ is defined and finite everywhere, measurable, and zero except on a set of measure zero. So $\int g_1d\mu=\int g_2d\mu$. I would encourage you to think about $L_1$ as the setting in which Folland is working, although the presentation isn't explicit enough about this. Roughly, what's happening here is that we don't care about things happening on subsets of measure zero. By "things happening" on subsets of measure zero, we mean lack of definition/lack of finiteness/non-zero differences.
Now we have integrals defined on $L_1$, which is a set of equivalence classes. So Folland's replacement of $f$ at the beginning of his proof of the dominated convergence theorem is allowed, because it's the same equivalence class.
Working with the completion gets us partway there, but not all the way to $L_1$, because, as you say, functions which are undefined on a subset of a set of measure zero aren't measurable in either $(\mathcal{M},\mu)$ or $(\overline{\mathcal{M}},\overline{\mu})$. But, for example, if $B\in N\setminus \mathcal{M}$, $1_B$ is not $\mathcal{M}$ measurable but it is $\overline{\mathcal{M}}$ measurable.
How far does going to the completion get us? Suppose that $f$ is defined and finite on $D$, and $E:= X\setminus D\in N$. Then $E\in \overline{\mathcal{M}}$ and $\overline{\mu}(E)=0$. Let $h:E\to \mathbb{R}$ be any function. Define $$g(x)=\left\{\begin{array}{ll} f(x) & : x\in D \\ h(x) & : x\in E.\end{array}\right.$$ Then the $\sim$-equivalence class $[f]$ of $f$ in $M_1$ lies in $L_1$ iff $g\in M_0$, and in this case $g\in M_0\cap [f]$.